# Let us consider an cc by nc nd 2011 j m powers 294

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Let us consider an CC BY-NC-ND. 2011, J. M. Powers.

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294 CHAPTER 10. CYCLES T s v 2 /2 v 2 /2 q in q in C T w t w c Figure 10.11: Sketch of turbojet schematic and associated T s plane. T s C T v 2 /2 w c w t v 2 /2 q in q in q in q in Figure 10.12: Sketch of turbojet with afterburners schematic and associated T s plane. Air standard analysis : a common set of assumptions used for idealized cyclic devices. The air standard make many compromises in order to admit some very simple analysis tools to be used to make simple estimates for the performance of a variety of devices. Actual design calculations would have to remedy the many shortcomings. But it is useful for a framework of understanding. We take the air standard to entail The working fluid is air. This ignores any effect of the properties of the fuel or any other fluid which is mixed with the air. The working fluid is an ideal gas. We will often assume it is a CPIG, but sometimes not. We will ignore all details of the combustion process and treat it as a simple heat addition. CC BY-NC-ND. 2011, J. M. Powers.
10.2. BRAYTON 295 T s v 2 /2 v 2 /2 q in q in Figure 10.13: Sketch of ramjet schematic and associated T s plane. Often in cycle analysis, the formal sign convention is ignored. We take the following Turbine work: w t = h 3 h 4 . Here the sign convention is maintained. Compressor work: w c = h 2 h 1 . Here the sign convention is ignored. Heat addition: q in = h 3 h 2 . Here the sign convention is maintained. Heat rejection: q out = h 4 h 1 . Here the sign convention is ignored. The cycle efficiency is η = ( h 3 h 4 ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright turbine ( h 2 h 1 ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright compressor h 3 h 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright combustor . (10.76) Rearranging Eq. (10.76), we can also say η = 1 h 4 h 1 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright heat rejected h 3 h 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright heat added . (10.77) The back work ratio, bwr , is bwr = w c w t = h 2 h 1 h 3 h 4 . (10.78) Note the back work ratio will be seen to be much larger for gas phase power cycles than it was for vapor cycles. For Brayton cycles, we may see bwr 0 . 4. For Rankine cycles, we usually see bwr 0 . 01. CC BY-NC-ND. 2011, J. M. Powers.

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296 CHAPTER 10. CYCLES Now if we have a CPIG, we get Δ h = integraltext c P dT to reduce to Δ h = c P Δ T . So Eq. (10.76) reduces to η = c P ( T 3 T 4 ) c P ( T 2 T 1 ) c P ( T 3 T 2 ) , (10.79) = T 3 T 4 T 2 + T 1 T 3 T 2 , (10.80) = 1 T 4 T 1 T 3 T 2 , (10.81) = 1 T 1 T 2 parenleftBigg T 4 T 1 1 T 3 T 2 1 parenrightBigg (10.82) Now 1 2 is isentropic. Recall for a CPIG which is isentropic that T 2 /T 1 = ( P 2 /P 1 ) ( k 1) /k .
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• Spring '10
• Powers
• Thermodynamics, Heat engine, Carnot cycle, Gas turbine, Thermodynamic cycles, J. M. Powers

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